Let $F_{even}$ be the $E$ module with underlying vector space $C^{\otimes n}$, where the $(2i)$-th factor of $E$ acts on the left side of the $i$-th factor of $F_{even}$, and the $(2i+1)$-th factor of $E$ acts on the right side (i.e. left action of $C^{op}$) of the $i$-th factor of $F_{even}$. ($0 \le i \le n-1$ .)

Define $F_{odd}$ similarly (same underlying vector space $C^{\otimes n}$), but with the $(2i)$-th factor of $E$ acting on the left side of the $i$-th factor of $F_{odd}$, and the $(2i+1)$-th factor of $E$ acting on the right side of the
$(i+1)$-th (modulo $n$) factor of $F_{odd}$. (Note that "modulo $n$" means the $n$-th factor is the 0-th factor.)

I'm interested in the derived $Hom_E(F_{even}, F_{odd})$. We can think of this as living in a disk with its boundary divided into $2n$ segments, alternating between incoming and outgoing segments. For $n=1$ this is just the Hochschild cohomology of $C$. For $n=2$ it's what one might associate to a saddle bordism in a 1+1-dimensional TQFT, or rather, the space in which the TQFT invariant of the saddle bordism lives.

Does $Hom_E(F_{even}, F_{odd})$ have a name? Is it mentioned in the literature anywhere?

In what you just edited you have a direct product of algebras where it should be tensor product.
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Grétar AmazeenDec 9 '09 at 4:22

1

Now I'm confused. I don't think you need to involve the opposite algebra, since it makes perfect sense to let C act on C from the right. In general you would need to have the opposite algebra in there, but C is naturally a right C-module as well as a left C-module. Another thing, if you change it to C \otimes C^op \otimes... etc, then the definition of F_even doesn't really make sense, does it? Now you have the opposite algebra acting from the right, which is the same as the original algebra acting from the left, so for the i-th guy in C^{\otimes n} you have two copies of C acting on the left.
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Grétar AmazeenDec 9 '09 at 4:54